The rate of physisorption 

Physisorption is a reversible process which favors low temperature. 

According to Freundlich adsorption isotherm,

$\frac{\mathrm{x}}{\mathrm{m}}\propto {\mathrm{P}}^{\frac{1}{\mathrm{n}}}$

$\frac{\mathrm{x}}{\mathrm{m}}={\mathrm{KP}}^{\frac{1}{\mathrm{n}}}$

Where x is the mole fraction of adsorbate is, m is the mass of adsorbent and P is the pressure. (n>1)

When $\frac{1}{n}=1$, at low pressure

$\mathrm{x}/\mathrm{m}=\mathrm{KP} \mathrm{so}, \mathrm{x}/\mathrm{m}\propto \mathrm{P}$

And when $\frac{1}{\mathrm{n}}=0$, at high pressure

$\mathrm{x}/\mathrm{m}=\mathrm{K}$

So, it is independent of $\mathrm{P},$ pressure.

Therefore, option (B) is correct.